Mathematics — CBSE Class 10 board question

Given: Two concentric circles with centre O, radii R = 5 cm and r = 3 cm. AB is a chord of the larger circle touching the smaller circle at P.

To find: Length of AB.

Since AB is a tangent to the smaller circle at P, by Theorem 10.1:
$$OP \perp AB$$

In right triangle OPB:
$$OB^2 = OP^2 + PB^2$$
$$5^2 = 3^2 + PB^2$$
$$PB^2 = 25 - 9 = 16$$
$$PB = 4 \text{ cm}$$

Since the perpendicular from the centre bisects the chord, AP = PB = 4 cm.

$$AB = 2 \times PB = 2 \times 4 = \textbf{8 cm}$$

Source: Chapter 10, Section 10.2 (Theorem 10.1) and Exercise 10.2 Q.7

• The key property used: a tangent is perpendicular to the radius at the point of contact (Theorem 10.1), so OP ⊥ AB.
• Since OP ⊥ AB and O is the centre, OP bisects chord AB → AP = PB.
• Apply Pythagoras in △OPB with OB = 5 (radius of larger circle) and OP = 3 (radius of smaller circle).
• Examiners expect the diagram reasoning (perpendicularity + bisection) stated clearly before the calculation.